One side of the roof of a building slopes up at 38.0°. A roofer kicks a round, f

Question

One side of the roof of a building slopes up at 38.0°. A roofer kicks a round, flat rock that has been thrown onto the roof by a neighborhood child. The rock slides straight up the incline with an initial speed of 15.0 m/s. The coefficient of kinetic friction between the rock and the roof is 0.390. The rock slides 10.0 m up the roof to its peak. It crosses the ridge and goes into free fall, following a parabolic trajectory above the far side of the roof, with negligible air resistance. Determine the maximum height the rock reaches above the point where it was kicked.

Explanation / Answer

initial KE = ½mv² = ½ * M * (15m/s)² = 112.5m²/s² * M

work done by friction W = µmgcos*d = 0.390 * M * 9.8m/s² * cos38º * 10m = M*30m²/s²

increase in PE = mgh = M * 9.8m/s² * 10m * sin38º = M * 60.33m²/s²

Then the rock has KE = M(112.5 - 30 - 60.33)m²/s² = 22.1m²/s² * M = ½Mv²

so

v = 6.65 m/s

We have the launch angle and velocity:

max. height = (V·sin)² / (2g) = (6.65m/s * sin38)² / 19.6m/s² = 0.85 m

plus the height it gained along the roof = 10m * sin38º = 6.15 m

yields a total rise h = 6.15 m

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